Question:
What is the first step and the final product formed in the reaction of HBr with \( CH_3-CH(CH_3)-CH=CH_2 \)?
What is the first step and the final product formed in the reaction of HBr with \( CH_3-CH(CH_3)-CH=CH_2 \)?
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Always check for carbocation rearrangement (hydride/methyl shift) in alkene additions!
Updated On: Apr 22, 2026
- Protonation at more substituted carbon; \( CH_3-C(Br)(CH_3)-CH_2-CH_3 \)
- Protonation at less substituted carbon; \( CH_3-CH(CH_3)-CHBr-CH_3 \)
- Radical initiation; \( CH_3-CH(CH_3)-CH_2-CH_2Br \)
- Protonation followed by hydride shift; \( CH_3-C(Br)(CH_3)-CH_2-CH_3 \)
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The Correct Option is D
Solution and Explanation
Concept:
Addition of HBr to alkene follows Markovnikov's rule (in absence of peroxide).
Formation of most stable carbocation occurs first.
Step 1: Protonation of alkene.
\[ \mathrm{CH_3-CH(CH_3)-CH=CH_2 \xrightarrow{H^+}} \] \(H^+\) adds to the terminal carbon (less substituted) $\longrightarrow$ secondary carbocation forms at \(C_3\). \[ \mathrm{CH_3-CH(CH_3)-\overset{+}{C}H-CH_3} \]
Step 2: Carbocation rearrangement.
Hydride shift from C2 to C3 occurs to form a more stable tertiary carbocation: \[ CH_3-\overset{+}{C}(CH_3)-CH_2-CH_3 \]
Step 3: Attack of nucleophile.
\[ Br^- \text{ attacks tertiary carbocation} \]
Step 4: Final product.
Product formed is: \[ CH_3-C(Br)(CH_3)-CH_2-CH_3 \] (2-bromo-2-methylbutane, a tertiary bromo compound)
Step 1: Protonation of alkene.
\[ \mathrm{CH_3-CH(CH_3)-CH=CH_2 \xrightarrow{H^+}} \] \(H^+\) adds to the terminal carbon (less substituted) $\longrightarrow$ secondary carbocation forms at \(C_3\). \[ \mathrm{CH_3-CH(CH_3)-\overset{+}{C}H-CH_3} \]
Step 2: Carbocation rearrangement.
Hydride shift from C2 to C3 occurs to form a more stable tertiary carbocation: \[ CH_3-\overset{+}{C}(CH_3)-CH_2-CH_3 \]
Step 3: Attack of nucleophile.
\[ Br^- \text{ attacks tertiary carbocation} \]
Step 4: Final product.
Product formed is: \[ CH_3-C(Br)(CH_3)-CH_2-CH_3 \] (2-bromo-2-methylbutane, a tertiary bromo compound)
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